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AS Physics P1 MCQs Preparation Thread.

Oliveme

Oliveme

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leosco1995 said:
Q16)

Take moments about the point X on the wall. You know that the weight lies in the middle, so if you assume the distance XY to be 1m (or 'x' m or whatever you want), the weight would be 0.5m from X. And the anti-clockwise moment, F is given by F sin 30 and it is 1m from X. So form an equation and solve:

F sin 30 * 1 = 0.5 * 10
F = 10.0 N

Q17) ..This one is kind of tricky, I think. I'm not sure if my method is correct. Check out the attachment. I basically found the angle between the strings Y and Z and then resolved forces to find the component 'x'. I did 4 cos 53.1 and then multiplied by 2 because there is a force acting both upwards and downwards. It turns out I got the right answer, but again, don't know if this is the proper method.

Q31) It can't be A because a +ve particle is going away from the +ve charge in the centre, so the repulsive force is decreasing.
It can definitely be B because that's basically the opposite of what I said above.
It can't be C or D because the force in an electric field is the same. In D, the +ve particle will experience the same force at any point. So the answer should be B.
Click to expand...
Thank you so very much! :)
 
S

SalamH

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ppl pleez explain d followin
12/o/n/09-Q5,Q16,Q25
01/M/J/03-Q6,Q10,Q20,Q23,Q40
 
H

Hassi123

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Knight said:
View attachment 11932
Click to expand...
The answer would be C, because the angle with the tension is 53.1(tan rule, tan inv= 80/30), and the horizontal force would be 4cos 53.1 making it 2.4, but see there are two tensions with the same angle (XY and XZ) so the total force would be 2.4 x 2 =4.8 (they are the same because same tension and same angle)
 
X

xyz!

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Paparazzi478 said:
A laser emits light of wavelength 600 nm.
What is the distance, expressed as a number of wavelengths, travelled by the light in one
second?
A 5 × 108 B 5 × 1011 C 5 × 1014 D 5 × 1017

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf

the answer is C
can you tell me how?
Click to expand...
They have asked for no. of wavelengths per unit tym...so basically u hav to calculate the frequency
so dat is speed/wavelength
= (3*10^8)/(600*10^-9)
=5*10^14 ==> C
 
G

geek101

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iKhaled said:
need help in may june 2004

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_1.pdf
Q9
Q23
Q25
Q26
Click to expand...

Q9 - this question we are supposed to use projectiles...
use v^2 = u^2 + 2as for each of the s values and add em up, you get 19...and the closest answer is 20 so D it is.

Q23- diameter = 1 mm
so radius = 0.5 mm >> 5 x 10^-4 m
and area = pi r^2 therefore the cross sectional area of the wire is 7.85 x 10^-7 m square
now take the highest value of F for each range and calculate the stress for each value. If the stress calculated is more than the one mentioned, the wire snaps and you get your answer! which is C becuz 2000 N gives you a stress of 2.5 x 10^8

Q25 - k here, because the wave is moving you have to draw the wave obtained after some time, when you draw the wave you will see that P is going down the only option which says so is A....the answer!

Q26 - okky im not sure about this though, i linked it to the equation intensity = power/time. When the amplitude doubles the intensity increases by 4 and so power which is also teh value E here, increases by 4 so now its 4E. Now keeping the intensity constant, you are reducing S or area by 1/2 so the E has to be halfed too to keep the intensity the same, so it becomes 0.5E. 4E x 0.5E = 2E
and the answer is B.
 
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